What is a pseudotensor, really, and how to tell one? I have some trouble with the concept of a "pseudotensor". Wikipedia distinguishes between that and a tensor density (e.g., here, where both concepts are used simultaneously) while e.g. in Eric Weisstein's Mathworld they say that

A pseudotensor is sometimes also called a tensor density.

These are two manifestly incompatible statements, and if asked, I'm inclined to believe that Wikipedia's definition of a pseudotensor, other than a synonyme for a tensor density, can not be rigorously formulated, i.e., no such thing exists. The purpose of this question is to confirm or refute this claim, mathematically. Some details follow to justify my case. "Pseudo" is used in its "Wikipedia" meaning: changing sign under inversion, whatever inversion means (discussed below).
I would appreciate it if answerers read the question in full and consider my problem specifically (and using my "language", if possible), rather than going with "stock" definitions or examples. No "looking at something in a mirror" either unless you can put that into formulas and argue it's a passive or active transform of some kind. Also, bear in mind that the "usual" sources in your country / curriculum may not be readily available to me so I would greatly appreciate it if when citing them you could also quote the relevant part.
With this out of the way, I appreciate that all of these quantities are defined through formulas their elements in given coordinate systems undergo in transformations. However what most sources neglect is distinction between passive and active transforms, and, worse, between the orientation of a vector space and an orientation of its basis. For simplicity, let us only assume linear transforms of linear spaces; this should be WLOG for the purposes of the question as stated.
So, under a passive transform, one compares the decompositions of the same quantity in two bases $E = (e_1, \ldots, e_n)$ and $F = (f_1, \ldots, f_n)$, where $f_i = S^j_{\ i} e_j$, $S = (S^k_{\ l})_{k,l=1}^n \in GL(n)$. A $k$ times covariant and $l$ times contravariant tensor density $T$ of weight $w$ transforms, by the definition I know, as
$$\tilde T^{i_1,\ldots,i_l}_{j_1,\ldots,j_k} = (\det S)^w\ S^{m_1}_{\hphantom{m_1}j_1} \ldots S^{m_k}_{\hphantom{m_k}j_k}\ (S^{-1})^{i_1}_{\hphantom{i_1}n_1} \ldots (S^{-1})^{i_l}_{\hphantom{i_l}n_l}\ T\strut^{n_1, \ldots, n_l}_{m_1, \ldots, m_k}$$
where $T\strut^\ldots_\ldots$ and $\tilde T^\ldots_\ldots$ denote the components in $E$ and in $F$, respectively. (The sign convention of $w$ may differ.)
Since we are looking at the same object in a different systems of coordinates, there is no reason the object changes, only its representation. I've been told ([1], [2]) that a typical example of a pseudoscalar in $\mathbb{R}^3$ is a triple product, $(\vec a,\vec b,\vec c) = \vec a \cdot (\vec b \times \vec c)$ since it "changes sign" under "inversion". Consider, though: in my notation, the inversion is just another $GL(n)$ matrix $S^i_{\ j} = -\delta^i_{\ j}$, and the "coordinate-less" definition of the triple product is
$$s := (a,b,c) = \omega(a,b,c)$$
where $\omega$ is a certain 3-form associated with the space, known as its volume form. In a "right-handed" orthonormal basis $E$, that is,
$$e_i \cdot e_j = \delta_{ij}, \qquad \omega(e_1,e_2,e_3) = +1,$$
the components of $\omega$ are
$$\omega_{ijk} = \epsilon_{ijk}$$
and thus we can express $s$ as
$$s = \omega_{ijk} a^i b^j c^k = \epsilon_{ijk} a^i b^j c^k.$$
If we now pick a basis $F = (-e_1,-e_2,-e_3)$ (which is left-handed), the quantity $s$ transforms as
$$\tilde s = \tilde w_{ijk} \tilde a^i \tilde b^j \tilde c^k,$$
where all of $a,b,c$ are contravariant and $\omega$ is fully covariant, so
$$\tilde a^i = -a^i, \quad \tilde b^j = -b^j, \quad \tilde c^k = -c^k$$
and also
$$\tilde \omega_{ijk} = (-1)^3 \omega_{ijk} = -\epsilon_{ijk}$$
by the above relation. Thus
$$\tilde s = -\epsilon_{ijk} (-a^i) (-b^j) (-c^k) = +\epsilon_{ijk} a^i b^j c^k = +s,$$
as expected. Of course, just looking at the same quantity (be it $s$, $\omega$, a cross product or anything else) in a different basis there is no reason the quantity should change, only perhaps its numeric description – that's what a passive transform means.
I'm not opposed to believing that $s$ is as pseudoscalar, but I'm firm in saying that a passive transform can't manifest distinction from a proper scalar. This is because although the orientation of two bases can easily differ, the orientation of the space itself is its inherent property, unaffected by what basis we choose to decompose its elements, so there are no "orientation-changing" passive transforms. In other words, while the above result shows that $s$ is consistent with being a scalar (of density zero), it does not show it could not actually be a pseudoscalar, the coordinate inversion just did not test this because no change in orientation (of the space) actually took place.
(The very same reasoning works with no significant changes if $s$ is replaced by another typical example, the vector product, as defined using the Hodge star.)
This leaves active transforms that could possibly distinguish a pseudoscalar from a proper scalar. Well, defining (for comparison) $T = -\mathbb{I}$ and introducing new (actively transformed) vectors
$$A^i = T^i_j a_j = -\delta^i_{\ j} a_j = -a_i, \quad B^i = -b^i, \quad C^i = -c^i$$
and their triple product in the original vector space,
$$S = \omega(A,B,C) = \epsilon_{ijk} A^i B^j C^k = (-1)^3 \epsilon_{ijk} a^i b^j c^k,$$
we indeed obtain $-s$ (in accordance with the cited sources). But how is it justified to tell anything about the properties of $s$ when $S$ is a new quantity (only defined analogously)?
What's worse, if we take an "arbitrary" active transform given by a generic $T \in GL(n)$ and recompute $S = (A,B,C)$ for the images $A, B, C$, we get
$$S = \epsilon_{ijk}\ T^i_{\ p} T^j_{\ q} T^k_{\ r}\ a^p b^q c^r = \det T\ \epsilon_{pqr} a^p b^q c^r = \det T \cdot s,$$
so if we accept that $s$ is a pseudoscalar just because $S$ had a different sign for a particular choice of a $\det < 0$ matrix, we can see that the same reasoning would lead to $S$ changing its magnitude as well (making it actually behave like a scalar density, rather than a pseudoscalar, in terms of its transformation as expressed using $s$). One could argue that active transforms by $T \not\in O(3)$ are not physical, but then again – improper rotations can't be "reached" in real world any better than e.g. volume-nonpreserving ones.
 A: The volume form $\omega$ is a legitimate rank-3 tensor. Hence $\omega(a,b,c)$ is a legitimate scalar, and indeed $\tilde{s} = +s$. When people say that the scalar triple product is a pseudoscalar, they must define it by
$$ a\cdot(b \times c) := \epsilon_{ijk}a^i b^j c^k \,,$$
where $\epsilon_{ijk}$ is the Levi-Civita symbol – crucially, not the components of a legitimate rank-3 tensor.
Note that Mathworld is often flat-out wrong. In particular, in the first page you linked to, both instances of $\mathsf{A}$ should be replaced by $\mathsf{a}$.
A: You've already got there, but it seems that you're having trouble taking the consequences at face value. The thing that marks $s$ as a pseudoscalar is simply the transformation property
$$
S=\det(T)\cdot s,
$$
as you have derived it. However, you go a bit off-track when saying that this means that the magnitude of $S$ depends on the transformation, because the set of linear transformations $\mathrm{GL}(3)$ is just too big. The set of active transformations that can be said to implement equivalent physical situations is $\mathrm{O}(3)$, and there the magnitude of $S$ doesn't change since $|\det(T)| = 1$.
Moreover, $\mathrm{O}(3)$ splits into two distinct subsets, the subgroup $\mathrm{SO}(3)$ which can be reached with rigid rotations that can be physically implemented in the real world, and its coset $(-\mathbb I)\mathrm{SO}(3)$ which requires you to re-build a separate copy of your experiment with an opposite orientation. With that in mind, then:


*

*scalar quantities are those that are invariant under the full action of $\mathrm{SO}(3)$, while

*pseudoscalar quantities are those that change their sign under the action of any active transformation drawn from the coset $(-\mathbb I)\mathrm{SO}(3)$.


The point that the coset $(-\mathbb I)\mathrm{SO}(3)$ can't be "reached" in the real world is precisely why we speak about a "mirror-world" experiment, i.e. a completely separate copy of the experiment that cannot be smoothly transformed into the original one.
